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  • Need solution for RD Sharma maths class 12 chapter 21 Differential Equation exercise Fill in the blank question 19

Need solution for RD Sharma maths class 12 chapter 21 Differential Equation exercise Fill in the blank question 19

Answers (1)

Answer:

 y=\frac{1}{4}x^{2}+Cx^{-2}

Hint:

 Substituting the values

Given:

 x\frac{\mathrm{d} y}{\mathrm{d} x}+2y=x^{2}

Solution:

On dividing the given differential equation by x, we get

\frac{\mathrm{d} y}{\mathrm{d} x}+2\frac{y}{x}=x

 So, the differential equation is a linear differential equation, and we can observe that the given differential equation is in the form

\frac{\mathrm{d} y}{\mathrm{d} x}+Py=Q

where,

P=\frac{2}{x} \, and\, Q=x

Here we get the integrating factor as

I.F=e^{\int Pdx}

By substituting the values,

=e^{\int \frac{2}{x}dx}

We get,

\begin{aligned} &=e^{2log\, x} \\ &=e^{log\, x^{2}} \\ &=x^{2} \end{aligned}

Here, the general solution can be written as

\begin{aligned} &Y(I.F)=\int Q(I.F)dx+C \end{aligned}

Substituting the values we get

\begin{aligned} &y(x^{2})=\int x(x)^{2}dx+C \\ &x^{2}y=\frac{x^{4}}{4}+C \end{aligned}

Dividing the entire equation by x2

\begin{aligned} &y=\frac{x^{2}}{4}+\frac{C}{x^{2}} \\ \Rightarrow &y=\frac{1}{4}x^{2}+Cx^{-2} \end{aligned}

So, the answer is

\begin{aligned} &y=\frac{1}{4}x^{2}+Cx^{-2} \end{aligned}

Posted by

Gurleen Kaur

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